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2-D vector analysis help

  1. Feb 21, 2014 #1
    1. The problem statement, all variables and given/known data

    http://imgur.com/wNusHOw

    2. Relevant equations

    I have the solutions and how they did it. THey took the deriv for the velocity vector, and then using t=0 and t=14, they found e=3.5 and f=-0.125

    3. The attempt at a solution

    I understand the math, but I don't understand why this is correct
    1) why did they take the velocity vector?
    2) if I plug e and f back into the position vector, and using t=14 i should get θ=0, ie, j-component is 0. But I don't. So what am I doing wrong?
     
  2. jcsd
  3. Feb 21, 2014 #2

    tiny-tim

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    welcome to pf! :smile:

    hi princeton_wu! welcome to pf! :smile:
    because the graph gives you the direction of the velocity vector
    yes you do …

    j-component = e + 2ft = 3.5 - 3.5 = 0 :wink:
     
  4. Feb 21, 2014 #3
    i mean the r vector; shouldn't the J-component of the r-vector be 0 @ t=14? this way, the angle @ t=14 would be 0.
     
  5. Feb 21, 2014 #4

    tiny-tim

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    i'm not following your reasoning :confused:

    the graph shows that vj = 0 (because θ = 0) at t = 14, it says nothing about r :smile:
     
  6. Feb 21, 2014 #5
    sorry, I'm confused too :tongue:

    If you use t=14 and plug it in the original position vector, shouldn't the J-component of the position vector be 0? Reasoning being that for θ to be 0, the J-component has to be 0?

    thanks for your patience tiny tim!
     
  7. Feb 22, 2014 #6

    tiny-tim

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    (just got up :zzz:)
    but θ (given in the graph) is stated to be the angle of the velocity vector …

    i don't understand what you think that has to do with the position vector :redface:
     
  8. Feb 22, 2014 #7
    I mulled over it last night and I finallly got it. My problem stemmed from the fact that I didn't realize that a t vs θ graph is a velocity graph). Thanks Tim! :smile:
     
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